Question

In a circle of radius 5 cm , AB and AC are two chords such that AB = AC = 6 cm . Find the length of the chord BC .

Answer 1

AB and AC are two equal chord of a circle, therefore the centre of the circle lies on the bisector of $\angle BAC$.

OA is the bisector of $\angle BAC$.

Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.

P divides BC in the ratio $6:6=1:1$.

P is mid-point of BC.

OP $\perp$ BC.

In $△$ ABP, by pythagoras theorem,

$A{B}^2=A{P}^2+B{P}^2$

$B{P}^2=36-A{P}^2$ ....(1)

In $△$ OBP, we have

$O{B}^2=O{P}^2+B{P}^2$

$5^2=(5-AP{)}^2+B{P}^2$

$B{P}^2=25-(5-AP{)}^2$ .....(2)

From 1 & 2, we get,

$36-A{P}^2=25-(5-AP{)}^2$

$36=10AP$

$AP=3.6cm$

Substitute in equation 1,

$B{P}^2=36-(3.6{)}^2=23.04$

$BP=4.8cm$

$BC=2\times 4.8=9.6cm$